Two resistors of resistances `R_(1)=100pm3` ohm and `R_(2)=200pm4` ohm are connected (a) series , (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation `R = R_(1)+R_(2)` and for (b) `(1)/(R') = (1)/(R_(1))+(1)/(R_(2))` and `(DeltaR')/(R'^(2)) = (DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2))`

Explanation

Equivalent resistance in series connection,
`R_(S)+R_(1)+R_(2)`
`=100+200`
`=300 Omega`
`rArr` Absolute error in series combination,
`DeltaR_(S)=DeltaR_(1)+DeltaR_(2)`
`=3+4`
`=7 Omega`
`rArr` Absolute error in series combination
`R_(S)+-DeltaR_(S)`
`=300Delta+-7 Omega`
(b) Absolute error in series combination,
`R_(P)=(R_(1)R_(2))/(R_(1)+R_(2))=(100xx200)/(100+200)=(200)/(3)=66.7 Omega`
and absolute error in parallel combination,
`(DeltaR_(p))/((R_(P))^(2))=(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2))`
`:.DeltaR_(P)=(R_(P))^(2)xx(DeltaR_(1))/(R_(1)^(2))+(R_(p))^(2)xx(DeltaR_(2))/(R_(2))^(2)`
`=(66.7)^(2)xx(3)/((100)^(2))+(66.7)^(2)xx(4)/((200)^(2))`
`:.DeltaR_(P)=1.334667+0.444889`
`=1.779556`
`~~1.8 Omega`
`rArr` absolute error in parallel combination
`R_(P)=R_(p)+-DeltaR_(p)`
`=66.7 Omega+-1.8 Omega`